trace-reverse linear field lagrangian

[michael879]

Ok I feel like there is a really simple answer to this but I've been trying to get this to work for days and I just can't. Here is the basic problem:

The "free-field" lagrangian for weak perturbations in the metric on a flat space is:

[tex]L = \frac{1}{4}(\partial^\sigma h_{\mu\nu}\partial_\sigma h^{\mu\nu} - \partial^\mu h \partial_\mu h)+\frac{1}{2}\partial_\mu h^{\mu\nu}(\partial_\nu h - \partial^\sigma h_{\nu\sigma})[/tex]

Solving for the E-L equations, and using the gauge [tex]\partial_\nu h^{\mu\nu}=\frac{1}{2}\partial^\mu h[/tex], I get the following equation:

[tex]\partial^\sigma\partial_\sigma h_{\mu\nu} - \frac{1}{2} \eta_{\mu\nu}\partial^\mu\partial_\mu h = 0[/tex]

and plugging in the trace-reverse of h, it is easy to get the final result:

[tex]\partial^\sigma\partial_\sigma \overline{h}_{\mu\nu} = 0[/tex]

 

So far this all agrees with my GR book (which doesn't take the lagrangian approach so its a good check). The problem I'm having is when I try to use the trace reverse from step 1. The lagrangian I get (I pre-set the same gauge as above to save myself some typing) is:

[tex]L = \frac{1}{4}\partial^\sigma \overline{h}_{\mu\nu}\partial_\sigma \overline{h}^{\mu\nu} - \frac{1}{8}\partial^\mu \overline{h} \partial_\mu \overline{h}[/tex]

However, this is EXACTLY the same lagrangian as the normal field (again, only after the gauge I mentioned has been set), and will therefore give the same EOM. So in the end I get:

[tex]\partial^\sigma\partial_\sigma \overline{h}_{\mu\nu} - \frac{1}{2} \eta_{\mu\nu}\partial^\sigma\partial_\sigma \overline{h} = 0[/tex]

Am I going crazy?? My logic seems sound but this is an OBVIOUS inconsistancy. I'd be tempted to just ignore the trace-reversed field but it makes a lot of things simpler and I'd rather use it...

 

 

*edit* sorry, if someone could tell me how to do latex here that would be great...

Edited May 4, 2011 by michael879

[timo]

[ math] and [ /math], excluding the blank, of course.